Oh, obviously. I just question whether this is a commonly known definition (at least among specialists) or you’re just going off on your own inventing your own customized definitions.
So for you, a finite number must be expressible in terms of a finite sum of rational numbers. So if it can’t be expressed as a finite sum or if it can’t be expressed with rational numbers, then it isn’t finite. If we’re gonna talk about this, we might as well give it a name. I’m just gonna throw one out there. How 'bout “semi-finite”?
^ By this definition, would (0.\dot3) be a semi-finite number? I wouldn’t say so since it can be expressed as 1/3 (a finite sum consisting of just one rational number). You could say the same of (0.\dot9) since it can be expressed as 1/3 x 3, but this gives you 1 which you dispute (which is a whole other can of worms we could debate). It seems pretty obvious that you would say an infinite sum of irrational numbers isn’t finite (although it wouldn’t necessarily be infinite either). What about a finite sum of irrational numbers? For most irrational numbers, if you sum them, you probably still get an irrational number, which I’m guessing you’d say is semi-finite. But what about (\pi) + (4 - (\pi))? 4 - (\pi) gives us 0.85840734641… I don’t have a mathematical proof that this is also an irrational number but my gut tells me it is. So you’d have a finite sum of two irrational numbers which equals 1, obviously a finite number. But obviously, 1 can also be expressed as a finite sum of rational numbers; 1 is a finite sum of rational numbers consisting of just one term (1). But what about (\pi) + 1. This equals 4.14159265358979… which is obviously also an irrational number. I don’t know of any way to express this as a sum of rational numbers, but I derived it with a sum of one irrational number and one rational number. ← Do you agree that it still counts as a semi-finite number?
Anyway, I always thought of finite numbers as just not infinite numbers. You seem to think of (0.\dot9) or (\pi) as not finite because, what, they don’t have clear boundaries? I mean, a number like 4 clearly starts at exactly 0 and ends at exactly 4. But if you think there exists an infinitesimal between (0.\dot9) and 1, but you can’t say “the 9s end exactly here”, then the boundaries of (0.\dot9) are not clear, at least at the end closer to 1. ← Is that why you don’t think of it as finite? But then what do you say about (0.\dot3) which can be expressed as 1/3? I would think you’d say it doesn’t have a clear boundary either, at least not at the side furthest from 0. Does this make it semi-finite? Or does the fact that it can be expressed as 1/3 make it finite?
This wouldn’t prove that B’ has more members than N. Once again, the rule is: if you can map all the members of N onto all the members of B, then you know there are just as many members in B as there are in N (infinite). You’re twisting it to say: if you can map all the members of N onto some members of B and find that there are some members of B left over, then you know there are more members in B than in N. ← That’s not the rule (though I understand why you think it can be restated as such). What your example proves is that there are just as many odd members in B as there are in N (infinite). But because the number of odd members is infinite, and because adding more members to an already infinite set is still infinite, then adding the even members of B to the mapping still gives you the same amount (infinite). So you can map all the natural numbers onto every odd member of B, and you can map all the natural numbers onto every member of B (odd and even), and yes, Magnus, I’m saying they will both be the same amount (infinite) (this is what I’ve been saying all along). You really, really, really should watch the vsauce video I posted. He goes exactly into this–how to map the naturals onto two infinite sets. Here it is again:
[youtube]http://www.youtube.com/watch?v=SrU9YDoXE88[/youtube]
^ Go to 5:00
Oh God.